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and all the equations, both internal and superficial, involving
a and B are satisfied by a = 0, B0, and therefore (App. C.)
require a = 0, ẞ=0. By means of (a), (b), and (c) the Fourier
solution is seen to be of the form

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and, because of (d), the coefficients A; are to be found so as to
make

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They are therefore [as we see by taking in § 77, (13) and (14),
such that (p − §) = (§), and putting p = 26] as follows:-

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crease of from edge

729. It is remarkable how very rapidly the whole disturb- Rapid deance represented by this result diminishes inwards from the disturbance edge where the disturbing traction is applied (compare § 586): inwards. also how very much more rapidly the second term diminishes than the first; and so on.

Thus as

€ = 2·71828, et = 4·801, €2·303 = 10, eTM = 23·141, €2′′ = 535·5,

==

VOL. II.

18

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be solved.

which proves most strikingly the concluding statement of § 647. Problems to 730. We regret that limits of space compel us to leave uninvestigated the torsion-flexure rigidities of a prism and the flexural rigidities of a plate of aeolotropic substance: and to still confine ourselves to isotropic substance when, in conclusion, we proceed to find the complete integrals of the equations [§ 697 (2)] of internal equilibrium for an infinite solid under the influence of any given forces, and the harmonic solutions suitable for problems regarding spheres and spherical shells, and solid and hollow circular cylinders (§ 738) under plane strain. The problem to be solved for the infinite solid is this:

General

problem of infinite solid:

solved for isotropic substance.

Let in (6) of § 698, X, Y, Z be any arbitrary functions whatever of (x, y, z), either discontinuous and vanishing in all points outside some finite closed surface, or continuous and vanishing at all infinitely distant points with sufficient convergency to make RD converge to 0 as D increases to oo, if R be the resultant of X, Y, Z for any point at distance D from origin. It is required to find a, ẞ, y satisfying those equations [(16) of § 698], subject to the condition of each vanishing for infinitely distant points (that is, for infinite values of x, y, or z).

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(b) This shows that if we imagine a mass distributed through space, with density p given by

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8 must be equal to its potential at (x, y, z). For [§ 491 (c)] if

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Subtracting this from (1) divided by (m + n), we have

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for all values of (x, y, z). Now the convergency of XD, YD, ZD to zero when D is infinite, clearly makes V=0 for all infinitely distant points. Hence if S be any closed surface round the origin of co-ordinates, everywhere infinitely distant from it, the function (8V) is zero for all points of it, and satisfies (3) for all points within it. Hence [App. A. (e)] we must have d = V. In other words, the fact that (1) holds for all points of space gives determinately

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where X, Y, Z' denote the values of X, Y, Z for any point.

(x', y', z').

(c) Modifying by integration by parts, and attending to the prescribed condition of convergences, according to which, when a' is infinite,

integrated.

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which for most purposes is more convenient than (4).

(d) On precisely the same plan as (b) we now integrate each of the three equations (6) of § 698 separately for a, ß, y respectively, and find

a=u+U, B=v + V, y = w + W..........................

(7)

where u, v, w, U, V, W denote the potentials at (x, y, z) of

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each through all space. Thus if 8", X", Y", Z" denote the values of 8, X, Y, Z for a point (x", y", z"), we find, for a,

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if in this we substitute for 8" its value by (6) we have a expressed by the sum of a sextuple integral and a triple integral, the latter being the U of (7); and similarly for ẞ and Y. These expressions may, however, be greatly simplified, since we shall see presently that each of the sextuple integrals may be reduced to a triple integral.

(e) As a particular case, let X, Y, Z be each constant throughout a spherical space having its centre at the origin and radius a, and zero everywhere else. This by (6) will make - 8 the sum of the products of X, Y, Z respectively into the corresponding component attractions of a uniform distribution of matter of density 1/4 (m + n) through this space. Hence [$ 491 (b)]

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3 (m + n) p3

- 1

3 (m + n)

(Xx+Yy + Zz) for points outside the spherical space,

(Xx+Yy+ Zz) for points within the spherical space.

Now we may divide u of (8) into two parts, u' and u", depending on the values of dd/dx within and without the spherical space respectively; so that we have,

(10).

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The solution of (11), being simply the potential due to a uniform

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Again, if in (12) of App. B. we put m=2, n=-3, and
V_=d8/dx, we have

-3

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Investigation of displacement.

since, for r > a, dô/dx is a spherical harmonic of order – 3. And r3 dồ/dx is [App. B. (13)] a solid harmonic of degree 2: hence if [dd/dx] denote, for any point within the spherical space, the same algebraic expression as dồ/dx by (10) for the external space, дав Габ

a3 dx

is a function which, for all the interior space, satisfies

the equation Vu = 0, and is equal to r' dô/dx for points infinitely

near the surface, outside and inside respectively. Hence

205 Габ
a3 dx

to spherical

for interior space, and r2 dô/dx for exterior space, constitute the Force applied potential of a distribution of matter of density 3d8/dx outside uniformly the spherical space and zero within, and, so far as yet tested, portion of any layer of matter whatever distributed over the separating homogenespherical surface. To find the surface density of this layer we first, for an exterior point infinitely near the surface, take

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and, for an interior point infinitely near the surface,

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infinite

ous solid.

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